January 2007 6679 Mechanics M3 Mark Scheme Question Number 1.
Marks
Scheme
(a) Maximum speed when accel. = 0 (o.e.)
B1 (1)
(b)
1 (30 − x ) = v dv (acceln = … + attempt to integrate) dx 12 dv v2 x2 ⎞ 1⎛ ⎜ ⎟ ( + c) : Use of v = 30 x − dx 2 12 ⎜⎝ 2 ⎟⎠
M1 ↓ M1 A1
Substituting x = 30, v = 10 and finding c (= 12.5), or limits
↓ M1
v 2 = 25 + 5 x − 121 x 2 (o.e.)
A1 (5)
(a) Allow “acceln > 0 for x < 30, acceln < 0 for x > 30” Also “accelerating for x < 30, decelerating for x > 30” But “acceln < 0 for x > 30” only is B0 (b) 1st M1 will be generous for wrong form of acceln (e.g. dv/dx)! 3rd M1 If use limits, they must use them in correct way with correct values Final A1. Have to accept any expression, but it must be for v2 explicitly (not 1/2v2), and if in separate terms, one can expect like terms to be collected. Hence answer in form as above, or e.g. 121 300 + 60 x − x 2 ; also 100 − 121 (30 − x) 2
(
)
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A 2.
a G
h Height of cone = Hence h =
3 4
a = 3a tan α
a
a 4 tan θ = 3 = ⇒ θ = 53.1° 3 4a 1st M1 (generous) allow any trig ratio to get height of cone (e.g. using sin) 3rd M1 For correct trig ratio on a suitable triangle to get θ or complement (even if they call the angle by another name – hence if they are aware or not that they are getting the required angle)
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M1 A1 ↓ M1
↓ M1 A1 (5)
3
(a)
(b)
E.P.E. =
1 3.6mg 2 1 3.6mg ⎛ a ⎞ x = ⎜ ⎟ 2 a ⎝3⎠ 2 a = 0.2 mga
2
⎛ 4a ⎞ Friction = µmg ⇒ work done by friction = µ mg ⎜ ⎟ ⎝ 3 ⎠ 1 (3 relevant terms) Work-energy: 2 m.2 ga = µ mgd + 0.2 mga Solving to find µ :
µ = 0.6
(b) 1st M1: allow for attempt to find work done by frictional force (i.e. not just finding friction). 2nd M1: “relevant” terms, i.e. energy or work terms! A1 f.t. on their work done by friction
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M1 A1 A1 (3) M1 A1 M1 A1√ ↓ M1 A1 (6)
4.
(a) Energy:
1 2
m.3ag − 12 mv 2 = mga(1 + cosθ )
M1 A1
v 2 = ag (1 − 2 cosθ ) (o.e.) (b)
A1 (3) M1 A1
v2 T + mg cosθ = m a
A1 cso (3)
Hence T = (1 − 3 cosθ )mg (*) (c)
Using T = 0 to find cos θ
M1
Hence height above A =
4 3
a
Accept 1.33a (but must have 3+ s.f.)
A1 (2)
v = ag 2
(d)
consider vert motion: sin 2 θ =
8 9
(o.e.)
1 3
(v sin θ )2 = 2 gh
f.t. using cosθ =
1 3
in v
2
(with v resolved)
(or θ = 70.53, sin θ = 0.943) and solve for h (as ka)
h= OR consider energy:
1 2
4 27
2
Sub for v, θ and solve for h
h=
4 27
a or 0.148a (awrt)
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M1 A1 ↓ M1 A1
a or 0.148a (awrt)
m(v cosθ ) + mgh = 12 mv 2
B1√
(3 non-zero terms)
M1 A1 ↓ M1 A1
Question Number 5.
Marks
Scheme
(a)
b T cosθ = mg
B1
↔ T + T sin θ = mrω 2
M1 A1
(3 terms)
B1
r = h tan θ 2 mg (1 + sin θ ) = mω h sin θ cosθ cos θ
(eliminate r)
g ⎛ 1 + sin θ ⎞ ω = ⎜ ⎟ (*) h ⎝ sin θ ⎠
(solve for ω
2
(b)
(c)
ω2 =
g⎛ 1 ⎞ + 1⎟ ⎜ h ⎝ sin θ ⎠
>
3g g ⎛ 1 + sin θ ⎞ = ⎜ ⎟ h h ⎝ sin θ ⎠
2g ( sin θ < 1) ⇒ h
⇒ sin θ =
ω>
2
)
2g (*) h
1 2
2 3 T cosθ = mg ⇒ T = mg or 1.15mg (awrt) 3 (a) Allow first B1 M1 A1 if assume different tensions (so next M1 is effectively for eliminating r and T. (b) M1 requires a valid attempt to derive an inequality for ω . (Hence putting sin θ = 1 immediately into expression of ω 2 [assuming this is the critical value] is M0.)
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↓ M1 ↓ M1 A1 (7) M1 A1 (2) M1 A1
↓ M1 A1 (4)
6.
(a) Moments: π
∫
2
∫
2
1
1
∫
2
1
xy 2 dx = V x or
y 2 dx =
xy 2 dx =
∫
1
2
xy 2 dx = x ∫ y 2 dx
2
1
9 7
7 ) 96
(either)
(=
3 ) 32
(both)
⇒ required dist =
)
H 3
2 ⎛1⎞ π⎜ ⎟ , 3 ⎝2⎠ ⎡ 1 ⎤ ⎢⎣= 12 ( ρ )π ⎥⎦
( ρ)
( ρ)
9 7
− 1 = m (*) 2 7
S
T
7π 96
H + S ⎡ 5 ⎤ ⎢⎣= 32 ( ρ )π ⎥⎦
19 m 16 ⎡ 1 ⎤ ⎛ 19 ⎞ ⎢⎣= 12 ( ρ )π ⎥⎦ ⎜⎝ 16 ⎟⎠ + ( ρ ) 29 m x = 30
Dist of CM from base Moments:
(=
2
1 ⎡ 1 ⎤ dx = ⎢− 2 ⎥ 3 4x ⎣ 8 x ⎦1
(b) Mass
M1
1
2
1
Solving to find x (=
2
1 ⎡ 1 ⎤ dx = ⎢− 4 3 4x ⎣ 12 x ⎥⎦ 1
2
∫
∫
5 m x 7 7π ⎛ 5 ⎞ ⎡ 5 ⎤ ⎜ ⎟ = ⎢ ( ρ )π ⎥ x 96 ⎝ 7 ⎠ ⎣ 32 ⎦ or 0.967 m (awrt)
Allow distances to be found from different base line if necessary
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M1 A1 A1
↓ M1 A1 cso (6)
B1, M1
B1 B1 M1 A1 A1 (7)
7.
(a)
A
T=
λ=
0.8
λ 0.8
(0.05) = 0.25g
M1
(0.8)(0.25g) = 39.2 (*) 0.05
A1 (2)
0.05
x 39.2 ( x + 0.05) 0.8 mg – T = ma
M1
T =
(b)
(3 term equn)
39.2 ( x + 0.05) = 0.25 &x& (or equivalent) 0.8 &x& = – 196 x 2π 2π π SHM with period = = s (*) ω 14 7
0.25g –
v = 14 √{(0.1) 2 − (0.05) 2 }
(c)
= 1.21(24…) ≈ 1.21 m s–1 (3 s.f.) Accept 7√3/10 1.21.. ( = 0.1237 s ) g Complete method for time T ′ from B to slack.
(d) Time T under gravity =
[ ↑ e.g.
π
+ t, where 0.05 = 0.1sin 14t 28 OR T ′ , where – 0.05 = 0.1 cos 14 T ′ ]
A1 A1
↓ M1 A1 cso (6) M1 A1√ A1 (3) B1√
M1 A1 A1
T'' = 0.1496s
A1
Total time = T + T ′ = 0.273 s (b)
M1
(5)
1st M1 must have extn as x + k with k ≠ 0 (but allow M1 if e.g. x + 0.15), or must justify later For last four marks, must be using &x& (not a)
(c) Using x = 0 is M0 (d) M1 – must be using distance for when string goes slack. Using x = – 0.1 (i.e. assumed end of the oscillation) is M0
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